IB Mathematics AHL 5.11 Definite and indefinite integration-AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Let \( u = 2x + 3 \), so \( du = 2 \, dx \), \( dx = \frac{du}{2} \)
\( \int \frac{8}{2x + 3} \, dx = \int \frac{8}{u} \cdot \frac{du}{2} = \int \frac{4}{u} \, du = 4 \ln |u| + C = 4 \ln |2x + 3| + C \)
Since \( x \geq 0 \) in part (b), \( 2x + 3 > 0 \), so:
\( 4 \ln (2x + 3) + C \)
Result: \( 4 \ln (2x + 3) + C \) [4]
(b)
Area = \( \int_0^6 \frac{8}{2x + 3} \, dx = \left[ 4 \ln (2x + 3) \right]_0^6 \)
At \( x = 6 \): \( 4 \ln (2 \cdot 6 + 3) = 4 \ln 15 \)
At \( x = 0 \): \( 4 \ln (2 \cdot 0 + 3) = 4 \ln 3 \)
Combine: \( 4 \ln 15 – 4 \ln 3 = 4 \ln \left( \frac{15}{3} \right) = 4 \ln 5 \)
Form \( a \ln b \): \( a = 4 \), \( b = 5 \), both in \( \mathbb{N} \)
Result: \( 4 \ln 5 \) [3]
▶️ Answer/Explanation
(a)(i)
Expand: \( \left( \frac{1}{u} + 1 \right)^2 = \frac{1}{u^2} + 2 \cdot \frac{1}{u} \cdot 1 + 1^2 = \frac{1}{u^2} + \frac{2}{u} + 1 \)
Result: \( \frac{1}{u^2} + \frac{2}{u} + 1 \) [2]
(a)(ii)
Using (a)(i): \( \left( \frac{1}{x+2} + 1 \right)^2 = \frac{1}{(x+2)^2} + \frac{2}{x+2} + 1 \)
Integrate: \( \int \left( \frac{1}{(x+2)^2} + \frac{2}{x+2} + 1 \right) \, dx = \int \frac{1}{(x+2)^2} \, dx + \int \frac{2}{x+2} \, dx + \int 1 \, dx \)
\( = -\frac{1}{x+2} + 2 \ln |x+2| + x + C \)
Since \( x \geq 0 \), \( x + 2 > 0 \): \( -\frac{1}{x+2} + 2 \ln (x+2) + x + C \)
Result: \( -\frac{1}{x+2} + 2 \ln (x+2) + x + C \) [2]
(b)
Volume: \( V = \pi \int_0^2 \left( \frac{1}{x+2} + 1 \right)^2 \, dx = \pi \left[ -\frac{1}{x+2} + 2 \ln (x+2) + x \right]_0^2 \)
At \( x = 2 \): \( -\frac{1}{4} + 2 \ln 4 + 2 = -\frac{1}{4} + 4 \ln 2 + 2 \)
At \( x = 0 \): \( -\frac{1}{2} + 2 \ln 2 \)
Combine: \( \pi \left[ \left( -\frac{1}{4} + 4 \ln 2 + 2 \right) – \left( -\frac{1}{2} + 2 \ln 2 \right) \right] = \pi \left( 2 \ln 2 + \frac{9}{4} \right) \)
Form: \( \frac{\pi}{4} (8 \ln 2 + 9) \), where \( a = 9 \), \( b = 8 \), \( c = 2 \)
Result: \( \frac{\pi}{4} (9 + 8 \ln 2) \) [3]